Exact distribution of the MLE of the quantile function for the exponential distribution

Question

Considering the pdf of the exponential distribution $$ f(x,\theta) = \theta e^{-\theta x} $$ with $x>0$, and parameter $\theta > 0$. It is straightforward to find that the (biased) maximum likelihood estimator of $\hat{\theta}$ equals

$$ \hat{\theta} = \frac{n}{\sum_{i = 1}^n x_i}. $$

Moreover, the quantile function of the distribution is equal to:

$$ q_p = - \frac{\ln (1 - p)}{ \theta} $$

Here, at last, comes the question: is it possible for me to write the MLE of the quantile distribution $\hat{q}_p$ as: $$ \hat{q}_p = - \frac{\ln(1-p)}{\hat{\theta}} ? $$ Moreover, how can I determine the exact distribution of the MLE of the quantile function? My intuition tells me that it should be normal, with mean equal to $q_p$ and variance equal to the inverse of the Fisher information, i.e.: $$ \sqrt{n}\hat{q}_p \to^d \mathcal{N}(q_p, I^{-1}) $$

Am I correct? Thanks in advance to everyone willing to argue.

Answer 1

is it possible for me to write the MLE of the quantile distribution $\hat{q}_p$ as

Yes, if you have some bijective transformation $\tau = \tau(\theta)$, then the MLE for $\tau$ is $\hat \tau = \tau (\hat \theta)$, where $\hat \theta$ is the MLE for $\theta$; this is called invariance property of MLE (the comment by @BGM points to a stronger version of it).

how can I determine the exact distribution of the MLE

Hint: $\hat{q}_p$ is proportional to $n/\hat \theta = \sum_{i=1}^n X_i\simeq \Gamma(n,\theta)$.